Re: ReplaceAll (/.)

*To*: mathgroup at smc.vnet.net*Subject*: [mg69202] Re: ReplaceAll (/.)*From*: p-valko at tamu.edu*Date*: Fri, 1 Sep 2006 18:40:59 -0400 (EDT)*References*: <ed9491$tu$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

I think you really want (a b) /. Times -> List and it does return {a,b} The pattern x_Times would pick "a part of an expression that - after plugging into the function Times - gives True". There is no such part of your expression so it is returned intact. Regards, P.V. Bruce Colletti wrote: > Re Mathematica 5.2.0.0. > > Why doesn't: > > (a b) /. x_Times -> List @@ x > > return {a,b}? My (flawed) reasoning is that a b has form Times[a,b], and so x matches to Times[a,b]. List@@x replaces Times with List to yield List[a,b], or {a,b}...or so it would seem. > > The explanation is somehow linked to the fact that: > > (a b) /. x_Times :> List @@ x > > returns {a,b}, but I don't see why this works but not the above. > > Although List@@(a,b) returns {a,b}, I want to use the above ReplaceAll function (this example here is taken from a larger problem). > > Thankx. > > Bruce